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Perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. A study is conducted to determine if the percent of women who receive financial aid in undergraduate school is different from the percent of men who receive financial aid in undergraduate school. A random sample of undergraduates revealed these results. At \(\alpha=0.01,\) is there significant evidence to reject the null hypothesis? $$ \begin{array}{lcc} & \text { Women } & \text { Men } \\ \hline \text { Sample size } & 250 & 300 \\ \text { Number receiving aid } & 200 & 180 \end{array} $$

Short Answer

Expert verified
Reject the null hypothesis; there is significant evidence of a difference in financial aid proportions between women and men at the 0.01 significance level.

Step by step solution

01

State the Hypotheses

In hypothesis testing, we start by stating the null hypothesis (usually denoted as \(H_0\)) and the alternative hypothesis (denoted as \(H_1\)). For this problem:- Null Hypothesis \(H_0\): The proportion of women receiving financial aid is equal to the proportion of men receiving financial aid, \(p_w = p_m\).- Alternative Hypothesis \(H_1\): The proportion of women receiving financial aid is different from the proportion of men receiving financial aid, \(p_w eq p_m\). The claim pertains to \(H_1\).
02

Find the Critical Value(s)

For a two-tailed test at the \(\alpha = 0.01\) significance level, the critical values can be found using a standard normal \(Z\)-table. We look for \(Z\) values that correspond to the upper and lower 0.005 (since \(0.01/2 = 0.005\)) tails of the distribution. These are approximately \(Z = -2.576\) and \(Z = 2.576\).
03

Compute the Test Value

First, calculate the sample proportions:- Proportion of women receiving aid: \(\hat{p}_w = \frac{200}{250} = 0.8\)- Proportion of men receiving aid: \(\hat{p}_m = \frac{180}{300} = 0.6\)Compute the pooled proportion:\(\hat{p} = \frac{200 + 180}{250 + 300} = \frac{380}{550} \approx 0.6909\)The test statistic \(Z\) is calculated as follows:\[Z = \frac{(\hat{p}_w - \hat{p}_m) - 0}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_w} + \frac{1}{n_m})}}\]Where \(n_w = 250\) and \(n_m = 300\).Substitute the values:\[Z = \frac{(0.8 - 0.6)}{\sqrt{0.6909(1 - 0.6909)(\frac{1}{250} + \frac{1}{300})}} \approx \frac{0.2}{0.054} \approx 3.704\]
04

Make the Decision

Compare the computed test statistic \(Z = 3.704\) to the critical values \(-2.576\) and \(2.576\). Since \(3.704\) is greater than \(2.576\), it falls into the rejection region. Thus, we reject the null hypothesis.
05

Summarize the Results

At \(\alpha = 0.01\), the test statistic is in the rejection region, providing significant evidence to conclude that there is a difference between the proportion of women and men receiving financial aid in undergraduate school. Thus, the null hypothesis is rejected in favor of the alternative hypothesis.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Critical Value
In hypothesis testing, the critical value is a key concept. It is the threshold beyond which we decide whether to reject or fail to reject the null hypothesis. Depending on the chosen significance level, commonly denoted by \(\alpha\), the critical value is determined from statistical tables, like the standard normal \(Z\)-table for normal distribution tests.

For a two-tailed test, you will find two critical values: one on the low end and one on the high end of the probability distribution. In this example, we have \(\alpha = 0.01\), meaning the test is conducted with a 1% risk of rejecting the null hypothesis wrongly. The critical values, in this case, are identified as \(Z = -2.576\) and \(Z = 2.576\). These are the points beyond which the test statistic must fall for us to conclude that the observed effect is statistically significant. By understanding critical values, one can accurately interpret whether the hypothesis test results support or oppose the null hypothesis.
Test Statistic
The test statistic is a numerical summary of the data that we use in a hypothesis test. It helps determine how far the sample data deviate from the null hypothesis. In this example, the test statistic is calculated using the formula for comparing two sample proportions: \[Z = \frac{(\hat{p}_w - \hat{p}_m)}{\sqrt{\hat{p}(1-\hat{p})(\frac{1}{n_w} + \frac{1}{n_m})}}\]Where \(\hat{p}_w\) and \(\hat{p}_m\) are the sample proportions of women and men receiving aid, respectively. We also use the pooled proportion \(\hat{p}\) which combines both samples to give an overall proportion.

A test statistic of \(Z = 3.704\) was determined. This value is instrumental as it allows us to compare our sample results to what is expected if the null hypothesis were true. Essentially, this figure represents how many standard deviations the observed effect is from the mean under the null hypothesis.
Null Hypothesis
The null hypothesis, denoted as \(H_0\), is a foundational element in statistical hypothesis testing. It represents the default position or the statement of no effect or no difference. In this context, it posits that there is no difference in the proportion of women and men receiving financial aid. Formally, it is expressed as \(p_w = p_m\).

The role of the null hypothesis is crucial, as hypothesis testing seeks to determine whether there is sufficient evidence to reject it. By doing so, statisticians maintain objectivity, as they assume the status quo holds until proven otherwise. The process of hypothesis testing hinges dramatically on whether the test data provide significant evidence against the null hypothesis.
Alternative Hypothesis
The alternative hypothesis, denoted as \(H_1\), is the statement that the researcher seeks to support. It is the antithesis of the null hypothesis. In this example, the alternative hypothesis claims that the proportion of women receiving financial aid is not equal to that of men – \(p_w eq p_m\).

Understanding the alternative hypothesis is essential because it guides the direction of the test. With a two-tailed test approach, the alternative hypothesis accommodates the potential for either an increase or decrease in the effect being measured. It highlights that the primary focus of the test is identifying if a statistically significant difference exists, warranting a reassessment of current presumptions. Successfully rejecting the null hypothesis bolsters the case for the alternative hypothesis.

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Most popular questions from this chapter

Perform the following steps. Assume that all variables are normally distributed. a. State the hypotheses and identify the claim. b. Find the critical value. c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. A metropolitan children's museum open year-round wants to see if the variance in daily attendance differs between the summer and winter months. Random samples of 30 days each were selected and showed that in the winter months, the sample mean daily attendance was 300 with a standard deviation of \(52,\) and the sample mean daily attendance for the summer months was 280 with a standard deviation of \(65 .\) At \(\alpha=0.05,\) can we conclude a difference in variances?

Perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. The percentages of ontime arrivals for major U.S. airlines range from 68.6 to \(91.1 .\) Two regional airlines were surveyed with the following results. At \(\alpha=0.01\), is there a difference in proportions? $$ \begin{array}{lcc} & \text { Airline A } & \text { Airline B } \\ \hline \text { No. of flights } & 300 & 250 \\ \text { No. of on-time flights } & 213 & 185 \end{array} $$

Perform these steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. In a random sample of 80 Americans, 44 wished that they were rich. In a random sample of 90 Europeans, 41 wished that they were rich. At \(\alpha=0.01,\) is there a difference in the proportions? Find the \(99 \%\) confidence interval for the difference of the two proportions.

Perform each of these steps. Assume that all variables are normally or approximately normally distributed a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. Improving Study Habits As an aid for improving students’ study habits, nine students were randomly selected to attend a seminar on the importance of education in life. The table shows the number of hours each student studied per week before and after the seminar. At \(\alpha=0.10\), did attending the seminar increase the number of hours the students studied per week? $$ \begin{array}{l|rrrrrrrrr} \text { Before } & 9 & 12 & 6 & 15 & 3 & 18 & 10 & 13 & 7 \\ \hline \text { After } & 9 & 17 & 9 & 20 & 2 & 21 & 15 & 22 & 6 \end{array} $$

Perform each of the following steps. a. State the hypotheses and identify the claim. b. Find the critical value(s). c. Compute the test value. d. Make the decision. e. Summarize the results. Use the traditional method of hypothesis testing unless otherwise specified. A researcher wishes to see if there is a difference between the mean number of hours per week that a family with no children participates in recreational activities and a family with children participates in recreational activities. She selects two random samples and the data are shown. At \(\alpha=0.10\), is there a difference between the means? $$ \begin{array}{lrcl} & \bar{X} & \sigma & n \\ \hline \text { No children } & 8.6 & 2.1 & 36 \\ \text { Children } & 10.6 & 2.7 & 36 \end{array} $$

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